Answer :
Given:
T2; T3
& T4
Since T2 , T3 & T4 are in AP
Then; 2(T3) = T2 + T4
i.e.
(n-1)(n-2)-6(n-1)+6=0
n2-3n+2-6n+6+6=0
n2-9n+14=0
(n-2)(n-7)=0
n= 2,7
n=2 rejected for term 3rd
So n=7
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